leetcode 【 Search in Rotated Sorted Array 】python 实现

题目：

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array

代码：oj测试通过 Runtime: 53 ms

 1 class Solution:
 2     # @param A, a list of integers
 3     # @param target, an integer to be searched
 4     # @return an integer
 5     def search(self, A, target):
 6         # none case & zero case
 7         if A is None or len(A)==0 :
 8             return -1
 9         # binary search
10         start = 0
11         end = len(A)-1
12         while start<=end :
13             # one element left case
14             if start == end :
15                 if A[start]==target :
16                     return start
17                 else:
18                     return -1
19             # two elements left case
20             if start+1 == end :
21                 if A[start]==target :
22                     return start 
23                 elif A[end]==target :
24                     return end
25                 else:
26                     return -1
27             # equal or more than three elements case
28             mid = (start+end)/2
29             if A[mid]==target :
30                 return mid
31             elif A[mid]>target:
32                 if A[start]>A[mid] and A[end]<A[mid]:
33                     start = mid+1
34                 elif A[start]<A[mid] and A[end]<A[mid]:
35                     if A[end]>=target:
36                         start = mid+1
37                     else:
38                         end = mid-1
39                 elif A[start]>A[mid] and A[end]>A[mid]:
40                     end = mid-1
41                 else:
42                     end = mid-1
43             else:
44                 if A[start]>A[mid] and A[end]<A[mid]:
45                     end = mid-1
46                 elif A[start]<A[mid] and A[end]<A[mid]:
47                     start = mid+1
48                 elif A[start]>A[mid] and A[end]>A[mid]:
49                     if A[end]>=target :
50                         start = mid+1
51                     else:
52                         end = mid-1
53                 else:
54                     start = mid+1
55         return -1

思路：

这个就是binary search的思路。

个人没想出来什么好的方法，硬着头皮硬写了一个暴力解决方法。

传统的binary search只需要判断A[mid]与target的大小就可以了；但这道题是rotated array，光判断A[mid]是不够的。

还需要判断A[start] A[end]与A[mid]的大小才能判断，target可能落在[start,mid]区间还是[mid,end]区间。

自己的代码实在有些繁琐丑陋，估计有些if else条件可以合并，后续会改进。